1000 patients currently suffering from a disease were selected to study the effectiveness of treatment of four types of medicines — A, B, C and D. These patients were first randomly assigned into two groups of equal size, called treatment group and control group. The patients in the control group were not treated with any of these medicines; instead they were given a dummy medicine, called placebo, containing only sugar and starch. The following information is known about the patients in the treatment group. a. A total of 250 patients were treated with type A medicine and a total of 210 patients were treated with type C medicine. b. 25 patients were treated with type A medicine only. 20 patients were treated with type C medicine only. 10 patients were treated with type D medicine only. c. 35 patients were treated with type A and type D medicines only. 20 patients were treated with type A and type B medicines only. 30 patients were treated with type A and type C medicines only. 20 patients were treated with type C and type D medicines only. d. 100 patients were treated with exactly three types of medicines. e. 40 patients were treated with medicines of types A, B and C, but not with medicines of type D. 20 patients were treated with medicines of types A, C and D, but not with medicines of type B. f. 50 patients were given all the four types of medicines. 75 patients were treated with exactly one type of medicine
*Answer can only contain numeric values
Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:How many patients were treated with medicine type B?
Correct Answer : 340
Explanation
1000 patients are equally distributed into two groups treatment group and control group. We have some information regarding the effectiveness of medicines A, B, C and D on the treatment group. Let us start filling the data give in the restrictions in a four sets Venn diagram.
75 patients were treated exactly one type of medicine.
∴ 25 + x + 20 + 10 = 75
⇒ x = 20
We have only one unknown in type-A medicine. 220 + b = 250 ⇒ 30
100 patients were treated with exactly three types of medicines. 40 + 20 + 30 + a = 100
⇒ a =10
Now, we have only one unknown in each of the type C and type-D.
⇒ p = 210 - 190 = 20
= 500 - 350 = 150
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*Answer can only contain numeric values
Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:The number of patients who were treated with medicine types B, C and D, but not type A was:
Correct Answer : 10
Explanation
1000 patients are equally distributed into two groups treatment group and control group. We have some information regarding the effectiveness of medicines A, B, C and D on the treatment group. Let us start filling the data give in the restrictions in a four sets Venn diagram.
75 patients were treated exactly one type of medicine.
∴ 25 + x + 20 + 10 = 75
⇒ x = 20
We have only one unknown in type-A medicine. 220 + b = 250 ⇒ 30
100 patients were treated with exactly three types of medicines. 40 + 20 + 30 + a = 100
⇒ a =10
Now, we have only one unknown in each of the type C and type-D.
⇒ p = 210 - 190 = 20
= 500 - 350 = 150
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*Answer can only contain numeric values
Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:How many patients were treated with medicine types B and D only?
Correct Answer : 150
Explanation
1000 patients are equally distributed into two groups treatment group and control group. We have some information regarding the effectiveness of medicines A, B, C and D on the treatment group. Let us start filling the data give in the restrictions in a four sets Venn diagram.
75 patients were treated exactly one type of medicine.
∴ 25 + x + 20 + 10 = 75
⇒ x = 20
We have only one unknown in type-A medicine. 220 + b = 250 ⇒ 30
100 patients were treated with exactly three types of medicines. 40 + 20 + 30 + a = 100
⇒ a =10
Now, we have only one unknown in each of the type C and type-D.
⇒ p = 210 - 190 = 20
= 500 - 350 = 150
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*Answer can only contain numeric values
Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:The number of patients who were treated with medicine type D was:
Correct Answer : 325
Explanation
1000 patients are equally distributed into two groups treatment group and control group. We have some information regarding the effectiveness of medicines A, B, C and D on the treatment group. Let us start filling the data give in the restrictions in a four sets Venn diagram.
75 patients were treated exactly one type of medicine.
∴ 25 + x + 20 + 10 = 75
⇒ x = 20
We have only one unknown in type-A medicine. 220 + b = 250 ⇒ 30
100 patients were treated with exactly three types of medicines. 40 + 20 + 30 + a = 100
⇒ a =10
Now, we have only one unknown in each of the type C and type-D.
⇒ p = 210 - 190 = 20
= 500 - 350 = 150
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Passage - 2
A survey of 600 schools in India was conducted to gather information about their online teaching learning processes (OTLP). The following four facilities were studied. F1: Own software for OTLP F2: Trained teachers for OTLP F3: Training materials for OTLP F4: All students having Laptops The following observations were summarized from the survey. 1. 80 schools did not have any of the four facilities – F1, F2, F3, F4. 2. 40 schools had all four facilities. 3. The number of schools with only F1, only F2, only F3, and only F4 was 25, 30, 26 and 20 respectively. 4. The number of schools with exactly three of the facilities was the same irrespective of which three were considered. 5. 313 schools had F2. 6. 26 schools had only F2 and F3 (but neither F1 nor F4). 7. Among the schools having F4, 24 had only F3, and 45 had only F2. 8. 162 schools had both F1 and F2. 9. The number of schools having F1 was the same as the number of schools having F4
Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:What was the total number of schools having exactly three of the four facilities?
Explanation
The given information can be represented in the following Venn diagram.
Given, F2 = (a + x + 40 + x) + (30 + 26 + x + 45) = 313
It is also given that F1 and F2 = a + x + 40 + x = 162.
Hence, 30 + 26 + x + 45 = 313 - 162 = 151
Hence, x = 151 - (30 + 26 + 45) = 50
The number of schools that have exactly three facilities = 4x = 200
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Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:What was the number of schools having facilities F2 and F4?
Explanation
The given information can be represented in the following Venn diagram.
The number of schools having facilities F2 and F4 = 40 + x + 45 + x = 185
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Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:What was the number of schools having only facilities F1 and F3?
Correct Answer : 42
Explanation
The given information can be represented in the following Venn diagram.
Only F1 and F3 = b
Given F1 = F4
25 + b + x + c + a + x + 40 + x
= 24 + 20 + x + 45 + 40 + x + x + c
Hence, a + b = 64
it is given that a + x + 40 + x = 162.
As x = 50, a = 22
Hence, only F1 and F3 = b = 64 - 22 = 42
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*Answer can only contain numeric values
Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:What was the number of schools having only facilities F1 and F4?
Correct Answer : 20
Explanation
The given information can be represented in the following Venn diagram.
Only F1 and F4 = c
Exactly 1 + Exactly 2 + Exactly 3 + Exactly 4 = 600 - 80 = 520
(25+ 30+ 26+ 20)+ Exactly 2 + 200 + 40 = 520
Hence, Exactly 2 = 179 = a + 24 + b + c + 26 + 45
As a = 22 and b = 42, c = only F1 and F4 = 20.
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Passage - 3
Ten musicians (A, B, C, D, E, F, G, H, I and J) are experts in at least one of the following three percussion instruments: tabla, mridangam, and ghatam. Among them, three are experts in tabla but not in mridangam or ghatam, another three are experts in mridangam but not in tabla or ghatam, and one is an expert in ghatam but not in tabla or mridangam. Further, two are experts in tabla and mridangam but not in ghatam, and one is an expert in tabla and ghatam but not in mridangam. The following facts are known about these ten musicians. 1. Both A and B are experts in mridangam, but only one of them is also an expert in tabla. 2. D is an expert in both tabla and ghatam. 3. Both F and G are experts in tabla, but only one of them is also an expert in mridangam. 4. Neither I nor J is an expert in tabla. 5. Neither H nor I is an expert in mridangam, but only one of them is an expert in ghatam.
Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:Who among the following is DEFINITELY an expert in tabla but not in either mridangam or ghatam?
Explanation
The given information can be represented in the following Venn diagram
From (1), one of A and B will be in region c and the other one in region f.
From (2), D is in region d.
From (3), one of F and G will be in region a and the other one in f.
From (4), either both I and J will be in region c or one in region b and the other in region c.
From (5), one of H and I will be in region b and the other in region a.
By combining (4) and (5), I will be in region b, H in region a and J in region c.
Thus, we get the following.
Region a = 3 (F/G, H, C/E), Region b = 1 (I), Region c = 3 (A/B, J, C/E)
Region d = 1 (D), Region f = 2 (B/A, G/F)
H definitely is an expert only in Tabla
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Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:Which of the following pairs CANNOT have any musician who is an expert in both tabla and mridangam but not in ghatam?
Explanation
The given information can be represented in the following Venn diagram
From (1), one of A and B will be in region c and the other one in region f.
From (2), D is in region d.
From (3), one of F and G will be in region a and the other one in f.
From (4), either both I and J will be in region c or one in region b and the other in region c.
From (5), one of H and I will be in region b and the other in region a.
By combining (4) and (5), I will be in region b, H in region a and J in region c.
Thus, we get the following.
Region a = 3 (F/G, H, C/E), Region b = 1 (I), Region c = 3 (A/B, J, C/E)
Region d = 1 (D), Region f = 2 (B/A, G/F)
One of A and B, one of G and F are experts in both tabla and mridangam but not ghatam. Three of the choices has at least one of the above four musicians. But one (C and E) does not have any one of these four. Hence, that is the answer
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Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:Who among the following is DEFINITELY an expert in mridangam but not in either tabla or ghatam?
Explanation
The given information can be represented in the following Venn diagram
From (1), one of A and B will be in region c and the other one in region f.
From (2), D is in region d.
From (3), one of F and G will be in region a and the other one in f.
From (4), either both I and J will be in region c or one in region b and the other in region c.
From (5), one of H and I will be in region b and the other in region a.
By combining (4) and (5), I will be in region b, H in region a and J in region c.
Thus, we get the following.
Region a = 3 (F/G, H, C/E), Region b = 1 (I), Region c = 3 (A/B, J, C/E)
Region d = 1 (D), Region f = 2 (B/A, G/F)
J definitely is an expert only in Mridangam
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Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:If C is an expert in mridangam and F is not, then which are the three musicians who are experts in tabla but not in either mridangam or ghatam?
Explanation
The given information can be represented in the following Venn diagram
From (1), one of A and B will be in region c and the other one in region f.
From (2), D is in region d.
From (3), one of F and G will be in region a and the other one in f.
From (4), either both I and J will be in region c or one in region b and the other in region c.
From (5), one of H and I will be in region b and the other in region a.
By combining (4) and (5), I will be in region b, H in region a and J in region c.
Thus, we get the following.
Region a = 3 (F/G, H, C/E), Region b = 1 (I), Region c = 3 (A/B, J, C/E)
Region d = 1 (D), Region f = 2 (B/A, G/F)
Given, C is an expert in mridangam but not F. It means F and E are experts only in tabla. Thus, E, F and H will be the experts in tabla but not in mridangam or ghatam.
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Passage - 4
Four institutes, A, B, C, and D, had contracts with four vendors W, X, Y, and Z during the ten calendar years from 2010 to 2019. The contracts were either multi-year contracts running for several consecutive years or single-year contracts. No institute had more than one contract with the same vendor. However, in a calendar year, an institute may have had contracts with multiple vendors, and a vendor may have had contracts with multiple institutes. It is known that over the decade, the institutes each got into two contracts with two of these vendors, and each vendor got into two contracts with two of these institutes. The following facts are also known about these contracts. I. Vendor Z had at least one contract in every year. II. Vendor X had one or more contracts in every year up to 2015, but no contract in any year after that. III. Vendor Y had contracts in 2010 and 2019. Vendor W had contracts only in 2012. IV. There were five contracts in 2012. V. There were exactly four multi-year contracts. Institute B had a 7-year contract, D had a 4- year contract, and A and C had one 3-year contract each. The other four contracts were single-year contracts. VI. Institute C had one or more contracts in 2012 but did not have any contract in 2011. VII. Institutes B and D each had exactly one contract in 2012. Institute D did not have any
contract in 2010.
Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:In which of the following years were there two or more contracts?
Explanation
Institutes A, B, C, D Vendors W, X, Y, Z
Contracts Awarded are multiyear Contracts (consecutive years) or single Year Contract.
No institute had more than one contract with the single vendor.
Each Institute → Two contracts → two vendors
Each Vendor → two contracts → two institutes
V. Exactly 4 multi year contract (A-3 years, B-7 years, C-3 years, D-4 years). Exactly 4 single year contracts. In total 8 contracts.
I. Vendor Z had at least one contract in ever year. This is only possible if he had both the contracts which are multi years contract. As 7 + 1 = 8 not possible. Only possibility is (7 years + 3 years) or (3 years + 7 years) = 10 years.
II. Vendor X → Six years 2010, 2011, 2012, 2013, 2014, 2015. To have contract for theses six years, only possibility is left with multi years contracts (4 years + 3 years or 3 years + 4 years) with 1 year overlapping contract. PI. Note (3 years + 3 years) is not possible for vendor X as one 3 years is already awarded to vendor Z{(7 years + 3 years ) or (3 years + 7 years) = 10 years}.
So all the 4 multiyear contracts are awarded to Z (3 years, 7 years) & X (3 years, 4 years). So we are left with 4 single year contracts.
III. Vendor Y had contracts in 2010 & 2019. Both of these contracts are single year contracts.
Vendor W had contracts (two contracts) only in year 2012. So both of these are single year contracts.
IV. There are 5 contracts in 2012.
VI. Institute C had one or more contracts in 2012, but no contract in 2011.
VII. Institute B & D each had exactly one contract in 2012. Institute D did not have any contract in 2010.
Out of 5 contracts in 2012 {B, D, W, W and C at least one contract). Hence C has exactly one contract in 2012.
For Vendor X, 4 year contract is with D. It Can vary from (2011-2014) or (2012-2015). No contract can be awarded to him after 2015 (point II). Only possibility for 3 year contract is with A only, as C do not have any contract in 2011. Therefore Vendor X has a 3 year contract with institute A from 2010-2012 & a 4 year contract with institute D from 2012 to 2015 (so as to have one or more contract from 2012 to 2015).
Vendor Z can be allotted contracts with only Institute B & C Only. As vendor Z has at least one contract in every year ( point I) , thus the only possibility left is first 7 year contract with institute B and then 3 year contract with institute C ( as institute C do not have any contract in 2011).
In 2012 there are 5 contracts. Three contracts are already assigned. Remaining two are single year contracts of W in 2012. Also Institute C has at least one contract in 2012. {No institute had more than one contract with the same vendor (initial condition)}. Hence both vendor W 1 years contract cannot be with Institute C. Hence Exactly one contract of vendor W is with institute C. Further, Point VII, Institutes B and D each had exactly one contract in 2012 which is already assigned. Hence second 1 year contract of vendor W is with institute A in 2012 (only possibility).
Institute A has allotted two contracts to 2 vendors {X & W}. Institute C had allotted contracts to two vendors {W & Z}. Hence Vendor Y can have a 1 year contract with only Institute B in 2010, (only possibility left). Hence vendor Y had 1 year contract in 2019 with Institute D (only possibility left).
In 2015 there were two contracts with vendor Z and X
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Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:In how many years during this period was there only one contract?
Explanation
Institutes A, B, C, D Vendors W, X, Y, Z
Contracts Awarded are multiyear Contracts (consecutive years) or single Year Contract.
No institute had more than one contract with the single vendor.
Each Institute → Two contracts → two vendors
Each Vendor → two contracts → two institutes
V. Exactly 4 multi year contract (A-3 years, B-7 years, C-3 years, D-4 years). Exactly 4 single year contracts. In total 8 contracts.
I. Vendor Z had at least one contract in ever year. This is only possible if he had both the contracts which are multi years contract. As 7 + 1 = 8 not possible. Only possibility is (7 years + 3 years) or (3 years + 7 years) = 10 years.
II. Vendor X → Six years 2010, 2011, 2012, 2013, 2014, 2015. To have contract for theses six years, only possibility is left with multi years contracts (4 years + 3 years or 3 years + 4 years) with 1 year overlapping contract. PI. Note (3 years + 3 years) is not possible for vendor X as one 3 years is already awarded to vendor Z{(7 years + 3 years ) or (3 years + 7 years) = 10 years}.
So all the 4 multiyear contracts are awarded to Z (3 years, 7 years) & X (3 years, 4 years). So we are left with 4 single year contracts.
III. Vendor Y had contracts in 2010 & 2019. Both of these contracts are single year contracts.
Vendor W had contracts (two contracts) only in year 2012. So both of these are single year contracts.
IV. There are 5 contracts in 2012.
VI. Institute C had one or more contracts in 2012, but no contract in 2011.
VII. Institute B & D each had exactly one contract in 2012. Institute D did not have any contract in 2010.
Out of 5 contracts in 2012 {B, D, W, W and C at least one contract). Hence C has exactly one contract in 2012.
For Vendor X, 4 year contract is with D. It Can vary from (2011-2014) or (2012-2015). No contract can be awarded to him after 2015 (point II). Only possibility for 3 year contract is with A only, as C do not have any contract in 2011. Therefore Vendor X has a 3 year contract with institute A from 2010-2012 & a 4 year contract with institute D from 2012 to 2015 (so as to have one or more contract from 2012 to 2015).
Vendor Z can be allotted contracts with only Institute B & C Only. As vendor Z has at least one contract in every year ( point I) , thus the only possibility left is first 7 year contract with institute B and then 3 year contract with institute C ( as institute C do not have any contract in 2011).
In 2012 there are 5 contracts. Three contracts are already assigned. Remaining two are single year contracts of W in 2012. Also Institute C has at least one contract in 2012. {No institute had more than one contract with the same vendor (initial condition)}. Hence both vendor W 1 years contract cannot be with Institute C. Hence Exactly one contract of vendor W is with institute C. Further, Point VII, Institutes B and D each had exactly one contract in 2012 which is already assigned. Hence second 1 year contract of vendor W is with institute A in 2012 (only possibility).
Institute A has allotted two contracts to 2 vendors {X & W}. Institute C had allotted contracts to two vendors {W & Z}. Hence Vendor Y can have a 1 year contract with only Institute B in 2010, (only possibility left). Hence vendor Y had 1 year contract in 2019 with Institute D (only possibility left).
In three years {in 2016 (Z), 2017 (Z), 2018 (Z)} there were only one contract
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Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:Which institutes had multiple contracts during the same year?
Explanation
Institutes A, B, C, D Vendors W, X, Y, Z
Contracts Awarded are multiyear Contracts (consecutive years) or single Year Contract.
No institute had more than one contract with the single vendor.
Each Institute → Two contracts → two vendors
Each Vendor → two contracts → two institutes
V. Exactly 4 multi year contract (A-3 years, B-7 years, C-3 years, D-4 years). Exactly 4 single year contracts. In total 8 contracts.
I. Vendor Z had at least one contract in ever year. This is only possible if he had both the contracts which are multi years contract. As 7 + 1 = 8 not possible. Only possibility is (7 years + 3 years) or (3 years + 7 years) = 10 years.
II. Vendor X → Six years 2010, 2011, 2012, 2013, 2014, 2015. To have contract for theses six years, only possibility is left with multi years contracts (4 years + 3 years or 3 years + 4 years) with 1 year overlapping contract. PI. Note (3 years + 3 years) is not possible for vendor X as one 3 years is already awarded to vendor Z{(7 years + 3 years ) or (3 years + 7 years) = 10 years}.
So all the 4 multiyear contracts are awarded to Z (3 years, 7 years) & X (3 years, 4 years). So we are left with 4 single year contracts.
III. Vendor Y had contracts in 2010 & 2019. Both of these contracts are single year contracts.
Vendor W had contracts (two contracts) only in year 2012. So both of these are single year contracts.
IV. There are 5 contracts in 2012.
VI. Institute C had one or more contracts in 2012, but no contract in 2011.
VII. Institute B & D each had exactly one contract in 2012. Institute D did not have any contract in 2010.
Out of 5 contracts in 2012 {B, D, W, W and C at least one contract). Hence C has exactly one contract in 2012.
For Vendor X, 4 year contract is with D. It Can vary from (2011-2014) or (2012-2015). No contract can be awarded to him after 2015 (point II). Only possibility for 3 year contract is with A only, as C do not have any contract in 2011. Therefore Vendor X has a 3 year contract with institute A from 2010-2012 & a 4 year contract with institute D from 2012 to 2015 (so as to have one or more contract from 2012 to 2015).
Vendor Z can be allotted contracts with only Institute B & C Only. As vendor Z has at least one contract in every year ( point I) , thus the only possibility left is first 7 year contract with institute B and then 3 year contract with institute C ( as institute C do not have any contract in 2011).
In 2012 there are 5 contracts. Three contracts are already assigned. Remaining two are single year contracts of W in 2012. Also Institute C has at least one contract in 2012. {No institute had more than one contract with the same vendor (initial condition)}. Hence both vendor W 1 years contract cannot be with Institute C. Hence Exactly one contract of vendor W is with institute C. Further, Point VII, Institutes B and D each had exactly one contract in 2012 which is already assigned. Hence second 1 year contract of vendor W is with institute A in 2012 (only possibility).
Institute A has allotted two contracts to 2 vendors {X & W}. Institute C had allotted contracts to two vendors {W & Z}. Hence Vendor Y can have a 1 year contract with only Institute B in 2010, (only possibility left). Hence vendor Y had 1 year contract in 2019 with Institute D (only possibility left).
A → (X & W) in 2012, B → (Z & Y) in 2010
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Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:Which of the following is true?
Explanation
Institutes A, B, C, D Vendors W, X, Y, Z
Contracts Awarded are multiyear Contracts (consecutive years) or single Year Contract.
No institute had more than one contract with the single vendor.
Each Institute → Two contracts → two vendors
Each Vendor → two contracts → two institutes
V. Exactly 4 multi year contract (A-3 years, B-7 years, C-3 years, D-4 years). Exactly 4 single year contracts. In total 8 contracts.
I. Vendor Z had at least one contract in ever year. This is only possible if he had both the contracts which are multi years contract. As 7 + 1 = 8 not possible. Only possibility is (7 years + 3 years) or (3 years + 7 years) = 10 years.
II. Vendor X → Six years 2010, 2011, 2012, 2013, 2014, 2015. To have contract for theses six years, only possibility is left with multi years contracts (4 years + 3 years or 3 years + 4 years) with 1 year overlapping contract. PI. Note (3 years + 3 years) is not possible for vendor X as one 3 years is already awarded to vendor Z{(7 years + 3 years) or (3 years + 7 years) = 10 years}.
So all the 4 multiyear contracts are awarded to Z (3 years, 7 years) & X (3 years, 4 years). So we are left with 4 single year contracts.
III. Vendor Y had contracts in 2010 & 2019. Both of these contracts are single year contracts.
Vendor W had contracts (two contracts) only in year 2012. So both of these are single year contracts.
IV. There are 5 contracts in 2012.
VI. Institute C had one or more contracts in 2012, but no contract in 2011.
VII. Institute B & D each had exactly one contract in 2012. Institute D did not have any contract in 2010.
Out of 5 contracts in 2012 {B, D, W, W and C at least one contract). Hence C has exactly one contract in 2012.
For Vendor X, 4 year contract is with D. It Can vary from (2011-2014) or (2012-2015). No contract can be awarded to him after 2015 (point II). Only possibility for 3 year contract is with A only, as C do not have any contract in 2011. Therefore Vendor X has a 3 year contract with institute A from 2010-2012 & a 4 year contract with institute D from 2012 to 2015 (so as to have one or more contract from 2012 to 2015).
Vendor Z can be allotted contracts with only Institute B & C Only. As vendor Z has at least one contract in every year ( point I) , thus the only possibility left is first 7 year contract with institute B and then 3 year contract with institute C ( as institute C do not have any contract in 2011).
In 2012 there are 5 contracts. Three contracts are already assigned. Remaining two are single year contracts of W in 2012. Also Institute C has at least one contract in 2012. {No institute had more than one contract with the same vendor (initial condition)}. Hence both vendor W 1 years contract cannot be with Institute C. Hence Exactly one contract of vendor W is with institute C. Further, Point VII, Institutes B and D each had exactly one contract in 2012 which is already assigned. Hence second 1 year contract of vendor W is with institute A in 2012 (only possibility).
Institute A has allotted two contracts to 2 vendors {X & W}. Institute C had allotted contracts to two vendors {W & Z}. Hence Vendor Y can have a 1 year contract with only Institute B in 2010, (only possibility left). Hence vendor Y had 1 year contract in 2019 with Institute D (only possibility left).
D had a contract with Y in 2019
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Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:What BEST can be concluded about the number of contracts in 2010?
Explanation
Institutes A, B, C, D Vendors W, X, Y, Z
Contracts Awarded are multiyear Contracts (consecutive years) or single Year Contract.
No institute had more than one contract with the single vendor.
Each Institute → Two contracts → two vendors
Each Vendor → two contracts → two institutes
V. Exactly 4 multi year contract (A-3 years, B-7 years, C-3 years, D-4 years). Exactly 4 single year contracts. In total 8 contracts.
I. Vendor Z had at least one contract in ever year. This is only possible if he had both the contracts which are multi years contract. As 7 + 1 = 8 not possible. Only possibility is (7 years + 3 years) or (3 years + 7 years) = 10 years.
II. Vendor X → Six years 2010, 2011, 2012, 2013, 2014, 2015. To have contract for theses six years, only possibility is left with multi years contracts (4 years + 3 years or 3 years + 4 years) with 1 year overlapping contract. PI. Note (3 years + 3 years) is not possible for vendor X as one 3 years is already awarded to vendor Z{(7 years + 3 years) or (3 years + 7 years) = 10 years}.
So all the 4 multiyear contracts are awarded to Z (3 years, 7 years) & X (3 years, 4 years). So we are left with 4 single year contracts.
III. Vendor Y had contracts in 2010 & 2019. Both of these contracts are single year contracts.
Vendor W had contracts (two contracts) only in year 2012. So both of these are single year contracts.
IV. There are 5 contracts in 2012.
VI. Institute C had one or more contracts in 2012, but no contract in 2011.
VII. Institute B & D each had exactly one contract in 2012. Institute D did not have any contract in 2010.
Out of 5 contracts in 2012 {B, D, W, W and C at least one contract). Hence C has exactly one contract in 2012.
For Vendor X, 4 year contract is with D. It Can vary from (2011-2014) or (2012-2015). No contract can be awarded to him after 2015 (point II). Only possibility for 3 year contract is with A only, as C do not have any contract in 2011. Therefore Vendor X has a 3 year contract with institute A from 2010-2012 & a 4 year contract with institute D from 2012 to 2015 (so as to have one or more contract from 2012 to 2015).
Vendor Z can be allotted contracts with only Institute B & C Only. As vendor Z has at least one contract in every year ( point I) , thus the only possibility left is first 7 year contract with institute B and then 3 year contract with institute C ( as institute C do not have any contract in 2011).
In 2012 there are 5 contracts. Three contracts are already assigned. Remaining two are single year contracts of W in 2012. Also Institute C has at least one contract in 2012. {No institute had more than one contract with the same vendor (initial condition)}. Hence both vendor W 1 years contract cannot be with Institute C. Hence Exactly one contract of vendor W is with institute C. Further, Point VII, Institutes B and D each had exactly one contract in 2012 which is already assigned. Hence second 1 year contract of vendor W is with institute A in 2012 (only possibility).
Institute A has allotted two contracts to 2 vendors {X & W}. Institute C had allotted contracts to two vendors {W & Z}. Hence Vendor Y can have a 1 year contract with only Institute B in 2010, (only possibility left). Hence vendor Y had 1 year contract in 2019 with Institute D (only possibility left).
In 2010 {(A-X), (B-Z), (B-Y)} exactly three contracts.
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Question for CAT Previous Year Questions: Venn Diagrams & Set Theory
Try yourself:Which institutes and vendors had more than one contracts in any year?
Explanation
Institutes A, B, C, D Vendors W, X, Y, Z
Contracts Awarded are multiyear Contracts (consecutive years) or single Year Contract.
No institute had more than one contract with the single vendor.
Each Institute → Two contracts → two vendors
Each Vendor → two contracts → two institutes
V. Exactly 4 multi year contract (A-3 years, B-7 years, C-3 years, D-4 years). Exactly 4 single year contracts. In total 8 contracts.
I. Vendor Z had at least one contract in ever year. This is only possible if he had both the contracts which are multi years contract. As 7 + 1 = 8 not possible. Only possibility is (7 years + 3 years) or (3 years + 7 years) = 10 years.
II. Vendor X → Six years 2010, 2011, 2012, 2013, 2014, 2015. To have contract for theses six years, only possibility is left with multi years contracts (4 years + 3 years or 3 years + 4 years) with 1 year overlapping contract. PI. Note (3 years + 3 years) is not possible for vendor X as one 3 years is already awarded to vendor Z{(7 years + 3 years) or (3 years + 7 years) = 10 years}.
So all the 4 multiyear contracts are awarded to Z (3 years, 7 years) & X (3 years, 4 years). So we are left with 4 single year contracts.
III. Vendor Y had contracts in 2010 & 2019. Both of these contracts are single year contracts.
Vendor W had contracts (two contracts) only in year 2012. So both of these are single year contracts.
IV. There are 5 contracts in 2012.
VI. Institute C had one or more contracts in 2012, but no contract in 2011.
VII. Institute B & D each had exactly one contract in 2012. Institute D did not have any contract in 2010.
Out of 5 contracts in 2012 {B, D, W, W and C at least one contract). Hence C has exactly one contract in 2012.
For Vendor X, 4 year contract is with D. It Can vary from (2011-2014) or (2012-2015). No contract can be awarded to him after 2015 (point II). Only possibility for 3 year contract is with A only, as C do not have any contract in 2011. Therefore Vendor X has a 3 year contract with institute A from 2010-2012 & a 4 year contract with institute D from 2012 to 2015 (so as to have one or more contract from 2012 to 2015).
Vendor Z can be allotted contracts with only Institute B & C Only. As vendor Z has at least one contract in every year ( point I) , thus the only possibility left is first 7 year contract with institute B and then 3 year contract with institute C ( as institute C do not have any contract in 2011).
In 2012 there are 5 contracts. Three contracts are already assigned. Remaining two are single year contracts of W in 2012. Also Institute C has at least one contract in 2012. {No institute had more than one contract with the same vendor (initial condition)}. Hence both vendor W 1 years contract cannot be with Institute C. Hence Exactly one contract of vendor W is with institute C. Further, Point VII, Institutes B and D each had exactly one contract in 2012 which is already assigned. Hence second 1 year contract of vendor W is with institute A in 2012 (only possibility).
Institute A has allotted two contracts to 2 vendors {X & W}. Institute C had allotted contracts to two vendors {W & Z}. Hence Vendor Y can have a 1 year contract with only Institute B in 2010, (only possibility left). Hence vendor Y had 1 year contract in 2019 with Institute D (only possibility left).
Institutes { A → 2012, B → 2010}, Vendors { W → 2012 X → 2012}
FAQs on Venn Diagrams & Set Theory CAT Previous Year Questions with Answer PDF
1. What is a Venn diagram and how is it used in CAT exams?
A Venn diagram is a graphical representation of sets using circles or other shapes. It is used in CAT exams to visually represent the relationships between different sets or groups of objects or elements. Venn diagrams help in solving problems related to set theory, especially in analyzing intersections, unions, and complements of sets.
2. How can Venn diagrams be used to solve CAT exam questions on set theory?
Venn diagrams can be used to solve CAT exam questions on set theory by visually representing the given information and using logical deductions. By shading or labeling specific regions or areas within the diagram, the relationships between different sets can be better understood. This helps in identifying the correct answers or solutions for the given questions.
3. What are the key elements of a Venn diagram in the context of CAT exams?
The key elements of a Venn diagram in the context of CAT exams are the sets or groups being represented, the universal set (the total population or sample space), the individual elements or objects within the sets, and the overlapping or non-overlapping regions that represent the intersections, unions, or complements of the sets.
4. Can Venn diagrams be used to solve complex CAT exam questions involving multiple sets?
Yes, Venn diagrams can be used to solve complex CAT exam questions involving multiple sets. By using multiple overlapping circles or shapes, each representing a different set, the relationships between multiple sets can be visually analyzed. This allows for the identification of common elements, exclusive elements, and other set operations required to solve the given questions.
5. Are there any limitations or drawbacks of using Venn diagrams to solve CAT exam questions?
While Venn diagrams are a useful tool for solving CAT exam questions related to set theory, they have some limitations. Venn diagrams become more complex and difficult to interpret as the number of sets increases. Additionally, Venn diagrams may not be suitable for solving questions involving large or infinite sets. In such cases, alternate methods like algebraic representations or logical reasoning may be more effective.
